Is the Pythagorean closure of $\mathbb Q$ equal to the field of constructible numbers? A Pythagorean field is one in which every sum of two squares is again a square. $\mathbb Q$ is not Pythagorean, which is easy to see. I have read a theorem online which says that every field has a unique (up to isomorphism) Pythagorean closure. I haven't found the proof so I thought I should start with the most familiar field, $\mathbb Q$.  
I was thinking if it would be possible to somehow imagine or characterize the Pythagorean closure $\mathbb P$ of $\mathbb Q.$ I know that in the field $\mathbb E$ of all constructible numbers, every positive number is a square because it is possible to construct square roots of already constructed numbers (by drawing a certain right triangle and its altitude). So $\mathbb E$ must be Pythagorean. $(1)$ But is it equal to $\mathbb P?$ If it's not, then what is the $\mathbb P$-dimension of $\mathbb E$?
Surely, for any $q_1,q_2,\ldots,q_n\in \mathbb Q,$ we must have $\sqrt{q_1^2+q_2^2+\cdots + q_n^2}\in \mathbb P$ by a simple induction. $(2)$ Is this all we have to adjoin to $\mathbb Q$ to obtain $\mathbb P?$ It looks like it is but I haven't done anything with infinite extensions and I don't know how to handle this.
 A: No, the two fields are not the same:  It is a result of Hilbert that the Pythagorean field is the maximal totally real subfield of the field of constructible numbers.  So any constructible number which is not totally real (i.e., its minimal polynomial has complex roots) gives a non-Pythagorean Euclidean number.  An easy example is $\sqrt{1+\sqrt{2}}$ (with non-real conjugate $\sqrt{1-\sqrt{2}}$).  Generalizing this example, it's easy to see that $\mathbb{E}$ is infinite-dimensional over $\mathbb{P}$.  
I believe your second question is answer in the affirmative by using that $\mathbb{P}$ is the smallest subfield of the Euclidean numbers closed under the operation of $x\rightarrow \sqrt{1+x^2}$, and inducting on the number of such operations you'd have to apply.  I'd take a look at Roger Alperin's series of papers on trisections and origami for a good discussion of the fields in question (and others!).
A: I treat the concept of Pythagorean closure in these notes.  In fact, I speak briefly about a class of fields being "closable" and give the Pythagorean closure as an example of that.
Let $F$ be a field.  Here are two key (but easy) observations:
(CC1) Any algebraically closed field extension of $F$ is Pythagorean.
(CC2) If $\{K_i\}_{i \in I}$ is a family of Pythagorean field extensions of $F$ -- all contained in some large field $L$ -- then also $K = \bigcap_{i \in I} K_i$ is a Pythagorean field extension of $F$.
It follows that inside any algebraic closure $\overline{F}$ of $F$ there is a literally unique Pythagorean closure: the intersection of all Pythagorean algebraic extensions of $F$.  (When $F = \mathbb{Q}$, Cam McLeman has given a spot-on description of it.)
A: When I think of the question of how to prove that every field has a unique Pythagorean closure, the first thing that comes to mind is that I might start with the fact that every field has a unique algebraic closure.  Once you've got that, you wouldn't have to construct a field from scratch; you'd just look at the intersection of all subfields of the algebraic closure that include the field you started with and are closed under  $(a,b)\mapsto\sqrt{a^2+b^2}$.  The algebraic closure is closed under that operation, so you'd get something.
Then you'd have the problem of proving uniqueness.  I expect that means unique up to isomorphism.  So could there be distinct but isomorphic subfields of the algebraic closure that are Pythagorean closures of the one you started with?  I suspect not.
